\newproblem{lay:1_9_34}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.9.34}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $S:\mathbb{R}^p\rightarrow \mathbb{R}^n$ and $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be linear transformations. Show that the mapping
	$\mathbf{x} \rightarrow T(S(\mathbf{x}))$ is a linear transformation from $\mathbb{R}^p$ to $\mathbb{R}^m$.
}{
  % Solution
	We need to show that $\forall \mathbf{x}_1,\mathbf{x}_2\in\mathbb{R}$, $\forall c\in\mathbb{R}$
		\begin{itemize}
			\item $T(S(\mathbf{x}_1+\mathbf{x}_2))=T(S(\mathbf{x}_1))+T(S(\mathbf{x}_2))$ \\
						\begin{center}
							$\begin{array}{rcll}
								T(S(\mathbf{x}_1+\mathbf{x}_2))&=&T(S(\mathbf{x}_1)+S(\mathbf{x}_2)) & \text{Because S is linear} \\
								                               &=&T(S(\mathbf{x}_1))+T(S(\mathbf{x}_2)) & \text{Because T is linear} \\
							\end{array}$
						\end{center}
			\item $T(S(c\mathbf{x}_1))=cT(S(\mathbf{x}_1))$ \\
						\begin{center}
							$\begin{array}{rcll}
								T(S(c\mathbf{x}_1))&=&T(cS(\mathbf{x}_1)) & \text{Because S is linear} \\
								                   &=&cT(S(\mathbf{x}_1)) & \text{Because T is linear} \\
							\end{array}$
						\end{center}
		\end{itemize}
}
\useproblem{lay:1_9_34}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
